Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → *1(x, y)
*1(s(x), s(y)) → +1(x, y)
SUM(cons(x, l)) → +1(x, sum(l))
SUM(cons(x, l)) → SUM(l)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(s(x), s(y)) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(s(x), s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
*(x1, x2)  =  *(x1, x2)
sum(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
prod(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
nil > s1
nil > 0
cons2 > *2 > +2 > s1
cons2 > *2 > 0

Status:
s1: [1]
+2: [2,1]
0: []
*2: [1,2]
nil: []
cons2: [1,2]

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SUM(cons(x, l)) → SUM(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SUM(cons(x, l)) → SUM(l)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
cons(x1, x2)  =  cons(x1, x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
*(x1, x2)  =  *
sum(x1)  =  sum(x1)
nil  =  nil
prod(x1)  =  prod

Lexicographic path order with status [LPO].
Precedence:
SUM1 > s
cons2 > sum1 > +2 > s
cons2 > sum1 > 0 > s
cons2 > prod > * > 0 > s
nil > s

Status:
SUM1: [1]
cons2: [1,2]
+2: [2,1]
0: []
s: []
*: []
sum1: [1]
nil: []
prod: []

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

*1(s(x), s(y)) → *1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(s(x), s(y)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x2)
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
*(x1, x2)  =  *(x1, x2)
sum(x1)  =  sum(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
prod(x1)  =  prod(x1)

Lexicographic path order with status [LPO].
Precedence:
*^11 > s1
sum1 > +2 > s1
sum1 > 0 > s1
nil > s1
cons2 > +2 > s1
prod1 > 0 > s1
prod1 > *2 > +2 > s1

Status:
*^11: [1]
s1: [1]
+2: [1,2]
0: []
*2: [1,2]
sum1: [1]
nil: []
cons2: [1,2]
prod1: [1]

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROD(cons(x, l)) → PROD(l)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROD(x1)  =  PROD(x1)
cons(x1, x2)  =  cons(x1, x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
*(x1, x2)  =  *
sum(x1)  =  sum(x1)
nil  =  nil
prod(x1)  =  prod

Lexicographic path order with status [LPO].
Precedence:
PROD1 > s
cons2 > sum1 > +2 > s
cons2 > sum1 > 0 > s
cons2 > prod > * > 0 > s
nil > s

Status:
PROD1: [1]
cons2: [1,2]
+2: [2,1]
0: []
s: []
*: []
sum1: [1]
nil: []
prod: []

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE